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[Mathqa]Re: Galois group of the rationals
Posted:
Mar 10, 2001 2:36 PM
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On Sat, 10 Mar 2001, Charles Matthews wrote:
> Gaetan Chenevier wrote:
>> Is there an example of such a group?
>> Yes, take the subfield of C generated by all the square roots of
integers,
>> call it K.
>
> OK, the final sentence of my posting on this was incorrect. Let me try to
> understand the claim. If V is the direct sum of countably many copies of
> the cyclic group of order 2, we can consider V as a vector space over the
> field with two elements. Its dual vector space V* is the cartesian product
> of countably many cyclic groups of order two. The double dual V** will be
> bigger than V, which is the Pontryagin dual of V*. That's because V* is
> uncountable, and, not very constructively, one can choose a basis and assign
> values of a linear functional on it. Therefore there are many subgroups
> of index 2 in V* that aren't closed in the product topology.
The Galois group you get by adding square roots of all the
positive integers to Q is a vector space of dimension 2^countable over Z/2Z.
You could find a basis for that, starting with the countable basis for the
vector space which has all but finitely many components = . Call that
basis B. Then you can fill that out to a basis B' for the whole vector space.
A basis for B is the automorphisms that send a square root of one prime
to its negative, and leave everything else as it is. You could define a
character on the Galois group to be 1 for elements of B, and 0 for all
other elements of B'. The kernel of this character is a subgroup H of
index 2.
Then suppose you have a subgroup of index 2 which is the fixed field of
Q(sqrt(n)), some positive n. The character which has this subgroup as its
kernel is 0 on some automorphism which sends the square root of some
prime not a factor of n to its negative. So, H isn't such a subgroup.
I looked at the algebraic completion of Z/pZ, and for the Galois group
of that over Z/pZ, subgroups of finite index do always correspond to
fields.
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